A328149
Numbers whose set of divisors contains a quadruple (x, y, z, w) satisfying x^3 + y^3 + z^3 = w^3.
Original entry on oeis.org
60, 72, 120, 144, 180, 216, 240, 288, 300, 360, 420, 432, 480, 504, 540, 576, 600, 648, 660, 720, 780, 792, 840, 864, 900, 936, 960, 1008, 1020, 1080, 1140, 1152, 1200, 1224, 1260, 1296, 1320, 1368, 1380, 1440, 1500, 1512, 1560, 1584, 1620, 1656, 1680, 1710
Offset: 1
120 is in the sequence because the set of divisors {1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120} contains the quadruples {3, 4, 5, 6} and {6, 8, 10, 12}. The first quadruple is primitive.
- Y. Perelman, Solutions to x^3 + y^3 + z^3 = u^3, Mathematics can be Fun, pp. 316-9 Mir Moscow 1985.
-
with(numtheory):
for n from 3 to 2000 do :
d:=divisors(n):n0:=nops(d):it:=0:
for i from 1 to n0-3 do:
for j from i+1 to n0-2 do :
for k from j+1 to n0-1 do:
for m from k+1 to n0 do:
if d[i]^3 + d[j]^3 + d[k]^3 = d[m]^3
then
it:=it+1:
else
fi:
od:
od:
od:
od:
if it>0 then
printf(`%d, `,n):
else fi:
od:
-
nq[n_] := If[ Mod[n, 6]>0, 0, Block[{t, u, v, c = 0, d = Divisors[n], m}, m = Length@ d; Do[ t = d[[i]]^3 + d[[j]]^3; Do[u = t + d[[h]]^3; If[u > n^3, Break[]]; If[ Mod[n^3, u] == 0 && IntegerQ[v = u^(1/3)] && Mod[n, v] == 0, c++], {h, j+1, m - 1}], {i, m-3}, {j, i+1, m - 2}]; c]]; Select[ Range@ 1026, nq[#] > 0 &] (* program from Giovanni Resta adapted for the sequence. See A330893 *)
-
isok(n) = {my(d=divisors(n), m); if (#d > 3, for (i=1, #d-3, for (j=i+1, #d-2, for (k=j+1, #d-1, if (ispower(d[i]^3+d[j]^3+d[k]^3, 3, &m) && !(n%m), return (1));););););} \\ Michel Marcus, Nov 15 2020
A337036
Numbers m such that the elements of all quadruples (x, y, z, w) satisfying x^3 + y^3 + z^3 = w^3 contained in the set of divisors of m are exactly the first k divisors of m for some k.
Original entry on oeis.org
720, 864, 1440, 1728, 2160, 2592, 2880, 3456, 4320, 5184, 5760, 6480, 6912, 7776, 8640, 10368, 11520, 12960, 13824, 15552, 17280, 19440, 20736, 23040, 23328, 25920, 27648, 31104, 34560, 38880, 41472, 46080, 46656, 51840, 55296, 58320, 62208, 69120, 69984, 77760
Offset: 1
2592 is in the sequence because the divisors are {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32, 36, 48, 54, 72, 81, 96, 108, 144, 162, 216, 288, 324, 432, 648, 864, 1296, 2592} and the elements of the 9 quadruples (x, y, z, w) satisfying x^3 + y^3 + z^3 = w^3 and belonging to the set of divisors of 2592: (1, 6, 8, 9), (2, 12, 16, 18), (3, 18, 24, 27), (4, 24, 32, 36), (6, 36, 48, 54), (9, 54, 72, 81), (12, 72, 96, 108), (18, 108, 144, 162) and (36, 216, 288, 324) are the first 25 divisors of 2592 with 25 = tau(2592) - 5 = 30 - 5.
- Y. Perelman, Solutions to x^3 + y^3 + z^3 = u^3, Mathematics can be Fun, pp. 316-9 Mir Moscow 1985.
-
with(numtheory):
for n from 6 by 6 to 200000 do :lst:={}:lst1:={}:it:=0:
d:=divisors(n):n0:=nops(d):
for i from 1 to n0-3 do:
for j from i+1 to n0-2 do:
for k from j+1 to n0-1 do:
for m from k+1 to n0 do:
if d[i]^3 + d[j]^3 + d[k]^3 = d[m]^3
then
it:=it+1:
lst:=lst union {d[i]} union {d[j]} union {d[k]} union {d[m]}:
else
fi:
od:
od:
od:
od:
n1:=nops(lst):
for l from 1 to n1 do:
lst1:= lst1 union {d[l]}:
od:
if lst=lst1 and lst<>{}
then
x:=tau(n)-n1:printf(`%d %d %d %d %d \n`,n,tau(n),n1,x,it):
else fi:
od:
-
isok(n) = {my(d=divisors(n), nb=0, s=[]); if (#d > 3, for (i=1, #d-3, for (j=i+1, #d-2, for (k=j+1, #d-1, if (ispower(d[i]^3+d[j]^3+d[k]^3, 3, &m) && !(n%m), s = concat(s, [d[i], d[j], d[k], m]));););); s = Set(s); if (#s, for (k=1, #s, if (s[k] != d[k], return (0));); return(1);););} \\ Michel Marcus, Nov 15 2020
A337098
Least k whose set of divisors contains exactly n quadruples (x, y, z, w) such that x^3 + y^3 + z^3 = w^3, or 0 if no such k exists.
Original entry on oeis.org
60, 120, 240, 432, 960, 360, 3840, 1728, 2592, 720, 1800, 2520, 161700, 1440, 6840, 9000, 2160, 2880, 168300, 5040, 41472, 5760, 1520820, 4320, 7200, 11520, 119700, 10080, 682080, 10800, 8640, 14400, 27360, 12960, 373248, 20160, 61560, 17280, 28800, 55440, 171000, 21600
Offset: 1
a(3) = 240 because the set of the divisors {1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 30, 40, 48, 60, 80, 120, 240} contains 3 quadruples {3, 4, 5, 6}, {6, 8, 10, 12} and {12, 16, 20, 24}. The first quadruple is primitive.
- Y. Perelman, Solutions to x^3 + y^3 + z^3 = u^3, Mathematics can be Fun, pp. 316-9 Mir Moscow 1985.
-
with(numtheory):divisors(240);
for n from 1 to 52 do :
ii:=0:
for q from 6 by 6 to 10^8 while(ii=0) do:
d:=divisors(q):n0:=nops(d):it:=0:
for i from 1 to n0-3 do:
for j from i+1 to n0-2 do :
for k from j+1 to n0-1 do:
for m from k+1 to n0 do:
if d[i]^3 + d[j]^3 + d[k]^3 = d[m]^3
then
it:=it+1:
else
fi:
od:
od:
od:
od:
if it = n
then
ii:=1: printf (`%d %d \n`,n,q):
else
fi:
od:
od:
-
With[{s = Array[Count[Subsets[Divisors[#], {4}]^3, ?(#1 + #2 + #3 == #4 & @@ # &)] &, 10^4]}, Rest@ Values[#][[1 ;; 1 + LengthWhile[Differences@ Keys@ #, # == 1 &] ]] &@ KeySort@ PositionIndex[s][[All, 1]]] (* _Michael De Vlieger, Sep 18 2020 *)
-
from itertools import combinations
from sympy import divisors
def A337098(n):
k = 1
while True:
if n == sum(1 for x in combinations((d**3 for d in divisors(k)),4) if sum(x[:-1]) == x[-1]):
return k
k += 1 # Chai Wah Wu, Sep 25 2020
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